Differential-Equations · Unit 4 · Video 4 · Interactive Practice

Homogeneous Equations $y' = F(y/x)$ : Unchanged by Zoom, Solved by $y = zx$

IKey Formulas

Formula	Name	What it says
$y' = F\!\left(\dfrac{y}{x}\right)$	Homogeneous equation	Every $x$ and $y$ on the right sits inside the one combination $y/x$ ; not the "homogeneous linear" equation $y' + p(x)y = 0$ , which shares only the name
$\dfrac{dy_1}{dx_1} = F\!\left(\dfrac{y_1}{x_1}\right)$	Zoom test	After the zoom $x = ax_1$ , $y = ay_1$ with $a > 0$ , the same equation in new names
$y = zx \;\Rightarrow\; y' = z'x + z$	Inverse substitution	The product rule gives $y'$ in $z$ , $z'$ and $x$ alone
$\dfrac{dz}{F(z) - z} = \dfrac{dx}{x}$	Separated form	Integrate, then put back $z = y/x$ ; each line $y = z_0 x$ with $F(z_0) = z_0$ is a solution the division loses

Key Insight: The equation contains $y'$ , so write the old variable in terms of the new one: $y = zx$ hands over $y' = z'x + z$ at once. The direct form $z = y/x$ gives $z' = \dfrac{xy' - y}{x^2}$ , still in the old variables; it reaches the same $x z' = F(z) - z$ only after $y' = F(z)$ and $y = zx$ are put in.

IIInvariant Under Zoom

Zoom slides each point along its ray from the origin, where $y/x$ is fixed, so $F(y/x)$ cannot change.

Equation Zoom factor a 2

For $x < 0$ , $\sqrt{x^2} = -x$ turns the second example into $y' = -\sqrt{1 + (y/x)^2}$ : still a function of $y/x$ , so it passes the zoom test there too.

IIIThe Four Steps on the First Example

The method solves $y' = \dfrac{x^2 y}{x^3 + y^3}$ , and the division in step 3 loses the line $y = 0$ .

Step 1

Step 1 — Recognise

y' = \frac{x^2 y}{x^3 + y^3} = \frac{y/x}{1 + (y/x)^3}

F(z) = \frac{z}{1 + z^3}

Step 2 — Substitute

y = zx, \quad y' = z'x + z

z'x + z = \frac{z}{1 + z^3}

x\,\frac{dz}{dx} = \frac{z}{1 + z^3} - z = -\frac{z^4}{1 + z^3}

Step 3 — Separate and integrate

\frac{1 + z^3}{z^4}\,dz = -\frac{dx}{x} \qquad (z \ne 0)

-\frac{1}{3z^3} + \ln|z| = -\ln|x| + c

Step 4 — Put back

z = \frac{y}{x}: \quad \ln|z| + \ln|x| = \ln|y|

\ln|y| = \frac{x^3}{3y^3} + c

x^3 = 3y^3\big(\ln|y| + C\big), \quad C = -c

IVQuiz Questions

Problem 1 · The Zoom Test

Given: the zoom $x = ax_1$ , $y = ay_1$ with $a > 0$ . Which equation comes back unchanged, and so is homogeneous?

✅ Correct! Top and bottom both have degree 2, so the zoom multiplies each by

a^2

and the factors cancel:

\dfrac{dy_1}{dx_1} = \dfrac{x_1^2 + y_1^2}{x_1 y_1}

❌ Not quite. The

y/x

term survives the zoom, but

y^2

becomes

a^2 y_1^2

, so the zoomed equation

\dfrac{dy_1}{dx_1} = \dfrac{y_1}{x_1} - a^2 y_1^2

is a different one. Containing

y/x

is not enough.

❌ Not quite. The numerator has degree 1 and the denominator degree 2, so the zoom leaves a factor

\dfrac{a}{a^2} = \dfrac{1}{a}

\dfrac{dy_1}{dx_1} = \dfrac{1}{a}\cdot\dfrac{x_1 + y_1}{x_1^2}

❌ Not quite. The constant

1

does not scale:

\dfrac{dy_1}{dx_1} = \dfrac{ay_1 + 1}{ax_1} = \dfrac{y_1}{x_1} + \dfrac{1}{ax_1}

, which still depends on

a

Show solution

Under the zoom $\dfrac{dy}{dx} = \dfrac{a\,dy_1}{a\,dx_1} = \dfrac{dy_1}{dx_1}$ , so only the right-hand side can change. Substitute $x = ax_1$ , $y = ay_1$ into (c):

\frac{(ax_1)^2 + (ay_1)^2}{(ax_1)(ay_1)} = \frac{a^2(x_1^2 + y_1^2)}{a^2 x_1 y_1} = \frac{x_1^2 + y_1^2}{x_1 y_1}

The equation comes back identical. Dividing top and bottom by $x^2$ shows its $F$ : $y' = \dfrac{1 + (y/x)^2}{y/x}$ , so $F(z) = \dfrac{1 + z^2}{z}$ .

The other three pick up $a$ : (a) becomes $\dfrac{y_1}{x_1} - a^2 y_1^2$ , (b) becomes $\dfrac{1}{a}\cdot\dfrac{x_1 + y_1}{x_1^2}$ , and (d) becomes $\dfrac{y_1}{x_1} + \dfrac{1}{ax_1}$ .

Problem 2 · Substituting for y′

Given: $x y' = \sqrt{x^2 + y^2}$ with $x > 0$ , that is $y' = \sqrt{1 + (y/x)^2}$ . Which separated equation does the substitution $y = zx$ produce?

✅ Correct!

y' = z'x + z

turns the equation into

x\,\dfrac{dz}{dx} = \sqrt{1 + z^2} - z

, whose right side depends on

z

alone.

❌ Not quite. This drops the

+z

from

y' = z'x + z

and treats

y'

z'x

. The product rule gives both terms, and moving that

z

to the right is what makes the denominator

F(z) - z

❌ Not quite. This uses

y' = z' + z

. The derivative of

zx

z'x + z

, and the

x

multiplying

z'

is what becomes

\dfrac{dx}{x}

on the right.

❌ Not quite. Moving

z

across the equals sign subtracts it:

x z' = \sqrt{1 + z^2} - z

, not

\sqrt{1 + z^2} + z

Show solution

Substitute both pieces, $y' = z'x + z$ on the left and $y/x = z$ on the right:

z'x + z = \sqrt{1 + z^2} \;\Longrightarrow\; x\,\frac{dz}{dx} = \sqrt{1 + z^2} - z

\frac{dz}{\sqrt{1 + z^2} - z} = \frac{dx}{x}

No line is lost by the division: $\sqrt{1 + z^2} > |z| \ge z$ , so $F(z) - z$ is never zero. Since $\big(\sqrt{1 + z^2} - z\big)\big(\sqrt{1 + z^2} + z\big) = 1$ , the left side is $\big(\sqrt{1 + z^2} + z\big)\,dz$ , and integrating gives

\tfrac{1}{2}\Big(z\sqrt{1 + z^2} + \ln\big(z + \sqrt{1 + z^2}\big)\Big) + \tfrac{1}{2}z^2 = \ln x + c,

with $z = y/x$ put back at the end.

Problem 3 · Solve It

Given: $y' = \dfrac{x^2 + y^2}{xy}$ with $x > 0$ . Find the general solution.

✅ Correct!

F(z) - z = \dfrac{1}{z}

, so

z\,dz = \dfrac{dx}{x}

\tfrac{1}{2}z^2 = \ln x + c

, and putting back

z = y/x

gives

y^2 = x^2(2\ln x + C)

❌ Not quite. That comes from

x z' = F(z)

, the substitution without its

+z

, and it solves

y' = \dfrac{x^2 + 2y^2}{xy}

instead. With

y' = z'x + z

the right side is

F(z) - z = \dfrac{1}{z}

❌ Not quite. Integrating

z\,dz

gives

\tfrac{1}{2}z^2

, so

z^2 = 2\ln x + 2c

: the

2

multiplies

\ln x

and cannot be absorbed into the constant.

❌ Not quite.

\tfrac{1}{2}z^2 = \ln x + c

is still in

z

. Putting back

z = y/x

gives

\dfrac{y^2}{x^2} = 2\ln x + C

, so the factor

x^2

belongs on the right.

Show solution

Step 1: divide top and bottom by $x^2$ :

F(z) = \frac{1 + z^2}{z} = \frac{1}{z} + z

Step 2: substitute $y = zx$ , $y' = z'x + z$ :

z'x + z = \frac{1}{z} + z \;\Longrightarrow\; x\,\frac{dz}{dx} = \frac{1}{z}

Step 3: separate and integrate:

z\,dz = \frac{dx}{x} \;\Longrightarrow\; \tfrac{1}{2}z^2 = \ln x + c

Step 4: put back $z = y/x$ , with $C = 2c$ :

\frac{y^2}{x^2} = 2\ln x + C \;\Longrightarrow\; y^2 = x^2\big(2\ln x + C\big)

Check: $2yy' = 2x(2\ln x + C) + 2x = \dfrac{2y^2}{x} + 2x$ , so $y' = \dfrac{y}{x} + \dfrac{x}{y} = \dfrac{x^2 + y^2}{xy}$ . No line is lost, because $F(z) - z = \dfrac{1}{z}$ is never zero.

Problem 4 · Lines the Division Loses

Given: $y' = \dfrac{y^2 + 2xy}{x^2}$ . Which straight lines $y = z_0 x$ are solutions?

✅ Correct! A line

y = z_0 x

has slope

z_0

, so it solves the equation exactly when

F(z_0) = z_0

z_0^2 + 2z_0 = z_0

, that is

z_0(z_0 + 1) = 0

❌ Not quite. The condition

z_0^2 + z_0 = 0

factors as

z_0(z_0 + 1) = 0

, so

z_0 = -1

is a root too, and

y = -x

checks:

y' = -1

and

\dfrac{x^2 - 2x^2}{x^2} = -1

❌ Not quite. That solves

F(z_0) = 0

, which would make the slope zero. A line

y = z_0 x

has slope

z_0

, so the condition is

F(z_0) = z_0

; on

y = -2x

the right side is

0

but the slope is

-2

❌ Not quite. On a line

y = z_0 x

the right side of a homogeneous equation is the constant

F(z_0)

, so the line is a solution whenever that constant equals its slope

z_0

, nonlinear or not.

Show solution

Divide top and bottom by $x^2$ : $y' = (y/x)^2 + 2(y/x)$ , so $F(z) = z^2 + 2z$ . On the line $y = z_0 x$ both sides are constant, $y' = z_0$ and $F(y/x) = F(z_0)$ :

F(z_0) = z_0 \iff z_0^2 + z_0 = 0 \iff z_0 = 0 \ \text{or}\ z_0 = -1

So the lines are $y = 0$ and $y = -x$ . They are exactly the zeros of $F(z) - z = z(z + 1)$ , the factor the separation step divides out:

\frac{dz}{z(z + 1)} = \frac{dx}{x} \;\Longrightarrow\; \ln\left|\frac{z}{z + 1}\right| = \ln|x| + c \;\Longrightarrow\; y = \frac{Kx^2}{1 - Kx}

$K = 0$ happens to bring back $y = 0$ , but no value of $K$ gives $y = -x$ : that solution survives only because it was found before dividing.

Solved: 0 / 4

Homogeneous Equations y′=F(y/x)y' = F(y/x)y′=F(y/x): Unchanged by Zoom, Solved by y=zxy = zxy=zx

IKey Formulas

IIInvariant Under Zoom

IIIThe Four Steps on the First Example

IVQuiz Questions

Homogeneous Equations $y' = F(y/x)$ : Unchanged by Zoom, Solved by $y = zx$